10.1 Comparing Two Proportions

Section 10.1Comparing Two Proportions

After this section, you should be able to…

DETERMINE whether the conditions for performing inference are met.

CONSTRUCT and INTERPRET a confidence interval to compare two proportions.

PERFORM a significance test to compare two proportions.

INTERPRET the results of inference procedures in a randomized experiment.

IntroductionSuppose we want to compare the proportions of

individuals with a certain characteristic in Population 1 and Population 2. Let’s call these parameters of interest p 1 and p 2. The ideal strategy is to take a separate random sample from each population and to compare the sample proportions with that characteristic.

What if we want to compare the effectiveness of Treatment 1 and Treatment 2 in a completely randomized experiment? This time, the parameters p 1 and p 2 that we want to compare are the true proportions of successful outcomes for each treatment.

The Sampling Distribution of a Difference Between Two

Proportions

The sampling distribution of a sample proportion has the following properties:

Shape Approximately Normal if np ≥ 10 and n(1 - p) ≥ 10

Center ˆ p p

Spread ˆ p p(1 p)

n if the sample is no more than 10% of the population

The Sampling Distribution of a Difference Between Two Proportions

To explore the sampling distribution of the difference between two proportions, let’s start with two populations having a known proportion of successes.

At School 1, 70% of students did their homework last night At School 2, 50% of students did their homework last night.

Suppose the counselor at School 1 takes an SRS of 100 students and records the sample proportion that did their homework.School 2’s counselor takes an SRS of 200 students and records the sample proportion that did their homework.

The Sampling Distribution of a Difference Between Two Proportions

Using Fathom software, we generated an SRS of 100 students from School 1 and a separate SRS of 200 students from School 2. The difference in sample proportions was then calculated and plotted. We repeated this process 1000 times. The results are below:

What do you notice about the shape, center, and spreadof the sampling distribution of ˆ p 1 ˆ p 2 ?

The Sampling Distribution of a Difference Between Two Proportions

Therefore, ˆ p 1 ˆ p 2

ˆ p 1 ˆ p 2

p1 p2

ˆ p 1 ˆ p 2

2 ˆ p 1

2 ˆ p 2

2

p1(1 p1)

n1

2

p2(1 p2)

n2

2

p1(1 p1)

n1

p2(1 p2)

n2

ˆ p 1 ˆ p 2 p1(1 p1)

n1

p2(1 p2)

n2

Choose an SRS of size n1 from Population 1 with proportion of successesp1 and an independent SRS of size n2 from Population 2 with proportion ofsuccesses p2.

Center The mean of the sampling distribution is p1 p2. That is,

the difference in sample proportions is an unbiased estimator of

the difference in population propotions.

Shape When n1p1, n1(1 p1), n2 p2 and n2(1 p2) are all at least 10, the

sampling distribution of ˆ p 1 ˆ p 2 is approximately Normal.

Spread The standard deviation of the sampling distribution of ˆ p 1 ˆ p 2 is

p1(1 p1)

n1

p2(1 p2)

n2

as long as each sample is no more than 10% of its population (10% condition).

Formula: Confidence Intervals for p 1 – p 2

statistic (critical value)(standard deviation of statistic)

( ˆ p 1 ˆ p 2) z *ˆ p 1(1 ˆ p 1)

n1

ˆ p 2(1 ˆ p 2)

n2

Two Proportion Z- Interval CI: Conditions

1) Random: Both sets of data should come from a well-designed random samples or randomized experiments.

3) Independent: Both sets of data must be independent. When sampling without replacement, the sample size n should be no more than 10% of the population size N (the 10% condition). Must check the condition for each separate sample.

Example: Teens and Adults on Social NetworksAs part of the Pew Internet and American Life Project, researchers conducted two surveys in late 2009. The first survey asked a random sample of 800 U.S. teens about their use of social media and the Internet. A second survey posed similar questions to a random sample of 2253 U.S. adults. In these two studies, 73% of teens and 47% of adults said that they use social-networking sites. Use these results to construct and interpret a 95% confidence interval for the difference between the proportion of all U.S. teens and adults who use social-networking sites.

Check Conditions:. Random The data come from two separate random samples Normal We check the counts of “successes” and “failures” and note the Normal condition is met since they are all at least 10:

Independent We clearly have two independent samples—one of teens and one of adults. Individual responses in the two samples also have to be independent. The researchers are sampling without replacement, so we check the 10% condition: there are at least 10(800) = 8000 U.S. teens and at least 10(2253) = 22,530 U.S. adults.

Teens and Adults on Social Networks

n1ˆ p 1 = 800(0.73) = 584 n1(1 ˆ p 1) 800(1 0.73) 216

n2ˆ p 2 = 2253(0.47) =1058.91 1059 n2(1 ˆ p 2) 2253(1 0.47) 1194.09 1194

State: We want to estimate the difference p1 – p2 at a 95% confidence level.

Teens and Adults on Social NetworksCalculations: Since the conditions are satisfied, we can construct a two-sample z interval for the difference p1 – p2.

Conclude: We are 95% confident that the interval from 0.223 to 0.297 captures the true difference in the proportion of all U.S. teens and adults who use social-networking sites. This interval suggests that more teens than adults in the United States engage in social networking by between 22.3 and 29.7 percentage points.

( ˆ p 1 ˆ p 2) z *ˆ p 1(1 ˆ p 1)

n1

ˆ p 2(1 ˆ p 2)

n2

(0.73 0.47) 1.960.73(0.27)

800

0.47(0.53)

22530.26 0.037

(0.223, 0.297)

An observed difference between two sample proportions can reflect an actual difference in the parameters, or it may just be due to chance variation in random sampling or random assignment. Significance tests help us decide which explanation makes more sense. The null hypothesis has the general form- H0: p1 - p2 = hypothesized value

We’ll restrict ourselves to situations in which the hypothesized difference is 0. Then the null hypothesis says that there is no difference between the two parameters:

H0: p1 - p2 = 0 or, alternatively, H0: p1 = p2

The alternative hypothesis says what kind of difference we expect.Ha: p1 - p2 > 0 or Ha: p1 - p2 < 0 or Ha: p1 - p2 ≠ 0

Significance Tests for p 1 – p 2

Significance Tests for p 1 – p 2

Significance Tests for p 1 – p 2

ˆ p C count of successes in both samples combinedcount of individuals in both samples combined

X1 X2

n1 n2

Use ˆ p C in place of both p1 and p2 in the expression for the denominator of the test statistic :

z ( ˆ p 1 ˆ p 2) 0

ˆ p C (1 ˆ p C )

n1

ˆ p C (1 ˆ p C )

n2

Why Do We Pool?“We have merged the data from the two samples to obtain what is called the "pooled" estimate of the standard deviation. We have done this not because it is more convenient (it isn't -- there's more calculation involved) nor because it reduces the measurement of variability (it doesn't always -- often the pooled estimate is larger*) but because it gives us the best estimate of the variability of the difference under our null hypothesis that the two sample proportions came from populations with the same proportion.”

For more details see: http://apcentral.collegeboard.com/apc/members/courses/teachers_corner/49013.html

Hungry ChildrenResearchers designed a survey to compare the proportions of children who come to school without eating breakfast in two low-income elementary schools. An SRS of 80 students from School 1 found that 19 had not eaten breakfast. At School 2, an SRS of 150 students included 26 who had not had breakfast. More than 1500 students attend each school. Do these data give convincing evidence of a difference in the population proportions? Carry out a significance test at the α = 0.05 level to support your answer.

Hungry ChildrenParameters & Hypotheses:

Our hypotheses areH0: p1 - p2 = 0Ha: p1 - p2 ≠ 0

where p1 = the true proportion of students at School 1 who did not eat breakfast, and p2 = the true proportion of students at School 2 who did not eat breakfast.

Assess Conditions: We should perform a two-sample z test for p1 – p2 if the conditions are satisfied. Random The data were produced using two simple random samples—of 80 students from School 1 and 150 students from School 2. Normal We check the counts of “successes” and “failures” and note the Normal condition is met since they are all at least 10:

Independent We clearly have two independent samples—one from each school. Individual responses in the two samples also have to be independent. The researchers are sampling without replacement, so we check the 10% condition: there are at least 10(80) = 800 students at School 1 and at least 10(150) = 1500 students at School 2.

n1ˆ p 1 =19, n1(1 ˆ p 1) 61, n2

ˆ p 2 = 26, n2(1 ˆ p 2) 124

Hungry Children

17.1

150)1957.01(1957.0

80)1957.01(1957.0

0)1733.02375.0(

)ˆ1(ˆ)ˆ1(ˆ

0)ˆˆ(=

21

21

npp

npp

ppz

CCCC

Name Test: Two-sample z test for the difference p1 – p2.

Test Statistic:

ˆ p C X1 X2

n1 n2

19 26

80 150

45

2300.1957

Obtain P-value: Using Table A or normalcdf, the desired P-value is2P(z ≥ 1.17) = 2(1 - 0.8790) = 0.2420.

Make a Decision: Since our P-value, 0.2420, is greater than the chosen significance level of α = 0.05,we fail to reject H 0.

Conclude: There is not sufficient evidence to conclude that the proportions of students at the two schools who didn’t eat breakfast are different.

Hungry Children

Cholesterol and Heart AttacksHigh levels of cholesterol in the blood are associated with higher risk of heart attacks. Will using a drug to lower blood cholesterol reduce heart attacks? The Helsinki Heart Study recruited middle-aged men with high cholesterol but no history of other serious medical problems to investigate this question. The volunteer subjects were assigned at random to one of two treatments: 2051 men took the drug gemfibrozil to reduce their cholesterol levels, and a control group of 2030 men took a placebo. During the next five years, 56 men in the gemfibrozil group and 84 men in the placebo group had heart attacks. Is the apparent benefit of gemfibrozil statistically significant? Perform an appropriate test to find out.

Cholesterol and Heart AttacksParameters & Hypotheses :

Our hypotheses areH0: p1 - p2 = 0 OR H0: p1 = p2

Ha: p1 - p2 < 0 Ha: p1 < p2

where p1 is the actual heart attack rate for middle-aged men like the ones in this study who take gemfibrozil, and p2 is the actual heart attack rate for middle-aged men like the ones in this study who take only a placebo. No significance level was specified, so we’ll use α = 0.01 to reduce the risk of making a Type I error (concluding that gemfibrozil reduces heart attack risk when it actually doesn’t).

Cholesterol and Heart AttacksAssess Conditions: We should perform a two-sample z test for p1 – p2 if the conditions are satisfied. Random The data come from two groups in a randomized experiment Normal The number of successes (heart attacks!) and failures in the two groups are 56, 1995, 84, and 1946. These are all at least 10, so the Normal condition is met. Independent Due to the random assignment, these two groups of men can be viewed as independent. Individual observations in each group should also be independent: knowing whether one subject has a heart attack gives no information about whether another subject does.

Cholesterol and Heart Attacks

Test statistic :

z =( ˆ p 1 ˆ p 2) 0

ˆ p C (1 ˆ p C )

n1

ˆ p C (1 ˆ p C )

n2

(0.0273 0.0414) 0

0.0343(1 0.0343)

2051

0.0343(1 0.0343)

2030

2.47

Name Test: Two-sample z test for the difference p1 – p2

Test Statistic (do Calculation):

ˆ p C X1 X2

n1 n2

56 84

2051 2030

140

40810.0343

Obtain P-value: Using Table A or normalcdf, the desired P-value is 0.0068

Cholesterol and Heart Attacks

Make a Decision: Since the P-value, 0.0068, is less than 0.01, the results are statistically significant at the α = 0.01 level.

State Conclusion: We can reject H0 and conclude that there is convincing evidence of a lower heart attack rate for middle-aged men like these who take gemfibrozil than for those who take only a placebo.

Standard Error

Confidence Intervals for p 1 – p 2

When data come from two random samples or two groups in a randomized

experiment, the statistic ˆ p 1 ˆ p 2 is our best guess for the value of p1 p2 . We

can use our familiar formula to calculate a confidence interval for p1 p2 :

statistic (critical value)(standard deviation of statistic)

When the Independent condition is met, the standard deviation of the statistic

ˆ p 1 ˆ p 2 is :

ˆ p 1 ˆ p 2

p1(1 p1)n1

p2(1 p2)

n2

Because we don't know the values of the parameters p1 and p2, we replace them

in the standard deviation formula with the sample proportions. The result is the

standard error of the statistic ˆ p 1 ˆ p 2 : ˆ p 1(1 ˆ p 1)

n1

ˆ p 2(1 ˆ p 2)

n2

If the Normal condition is met, we find the critical value z* for the given confidencelevel from the standard Normal curve. Our confidence interval for p1 – p2 is:

statistic (critical value)(standard deviation of statistic)

( ˆ p 1 ˆ p 2) z *ˆ p 1(1 ˆ p 1)

n1

ˆ p 2(1 ˆ p 2)

n2